By now you have had a lot of opportunity to practice
differentiating common functions.
And you have learned many techniques of Differentiation.
So for example, if I were to give you a function, let's call
this function capital F of X.
If it's a fairly simple function, you'll know
how to differentiate it.
And so by differentiating it, you'll be able to calculate its
derivative, which you'll remember we denote as DF DX. So
that's a process that you're very familiar with already.
Now in this unit we're going to refer to DF DX as little F of X.
So little F of X.
Is the derivative of big F of X. Let me write
that down little F of X.
Is the derivative.
As big banks.
And as I say, that's a process that you're very familiar with.
What we want to do now is try to work this process in reverse.
Work it backwards. In other words, we want to start with
little F. And try and come back this route.
And try and find the function or functions capital F which when
they are differentiated will give you little Earth. So we
want to carry out this process in reverse. We can think of this
as anti differentiation.
So we can think of this anti differentiation as
differentiation in reverse.
So big F of X we're going to call the anti derivative of
little F of X.
So by these two results in mind and try not to get confused with
a capital F in the lower case F little F is the derivative of
Big F. Because little F is DF DX.
And big F is the anti derivative of little laugh.
Now you may have already looked at a previous video called
integration as summation.
And in that video, the concept of an integral is defined in
terms of the sum of lots of rectangular areas under a curve.
To calculate an integral, you need to find the limit of a son.
And that's a very cumbersome and impractical process. What
we're going to learn about in this video is how to find
integrals, not by finding the limit of a sum, but instead
by using antiderivatives.
Let's start off with the example.
Suppose we start off with the function capital F of X.
Is equal to three X squared plus 7X.
Minus 2.
And what I'm going to do is something that you're already
very familiar with. We're going to differentiate it.
And if we differentiate it.
We get TF DX equals.
Turn by turn the derivative of three X squared is going to be.
2306 X.
And the derivative of Seven X is just going to be 7.
What about the minus two? Well, you remember that the derivative
of a constant is 0, so when we do this differentiation process,
the minus two disappears.
So our derivative DF DX is just six X +7.
So I'm going to write little F of X is 6 X +7.
And think about what will happen when we reverse the process when
we do anti differentiation.
Ask yourself what is an anti derivative of 6X plus 7?
Now we already have an answer to this question and anti
derivative of 6X plus 7.
Is 3 X squared plus 7X minus two, so the answer.
Is F of X is 3 X squared plus 7X minus two, but I'm afraid
that's not the whole story. We've already seen that when we
differentiate it, three X squared plus 7X minus two, the
minus two disappeared.
In other words, whatever number had been in here, whether that
minus two had been minus 8 or plus 10 or 0, whatever it would
still have disappeared. We've lost some information during
this process of Differentiation, and when we want to reverse it
when we want to start with the six, XX have Seven, and working
backwards, we've really no idea what that minus two might have
been. It could have been another
number. So this leads us to the conclusion that once we found an
antiderivative like this one, three X squared plus, 7X minus
2. Than any constant added onto this will still be an
anti derivative.
If F of X.
Is an anti derivative.
A little F of X.
Then so too.
Is F of X the anti derivative we've found plus any constant at
all we choose. See is what we call an arbitrary constant. Any
value of see any constant we can add on to the anti derivative
we've found and then we've got another antiderivative. So in
fact there are lots and lots of antiderivatives of 6X plus
Seven, just as another example, we could have had over here.
Three X squared plus 7X minus 8.
Or even just three X squared plus 7X where the constant term
was zero. So lots and lots of different antiderivatives for a
single term over here.
Let's do another example.
Suppose this time we look at.
A cubic function that suppose F of X is 4X cubed.
Minus Seven X squared.
Plus 12X
minus 4.
Again, you know how to differentiate this. You've
had a lot of practice differentiating functions
like this.
So the derivative DF DX is going to be.
Three 412 X squared.
The derivative of minus Seven X squared is going to
be minus 14X.
And the derivative of 12 X is just 12.
So think of our little F of X as being the function 12 X squared
minus 14X at 12.
Notice again the point that the minus four in the
differentiation process disappears.
So now ask the question, what's an anti derivative of this
function here little F.
Well, we've got one answer. It's on the page already. It's this
big F is an anti derivative of
Little F. But we've seen that any constant added on
here will still yield another antiderivative, so
we can write down lots of other ones, for example.
If we add on a constant.
And let's suppose we add on the constant 10 onto this one.
Then we'll get the function 4X cubed, minus Seven X squared.
Plus 12X plus six is also an anti derivative of Little F.
And the same goes by adding adding on any other constant at
all we choose. If we add on minus six to this.
Will get 4X cubed, minus Seven X squared plus 12X and adding on
minus six or just leave us this so that that is another
antiderivative. There's an infinite number of
antiderivatives of Little F.
Now in all the examples we've
looked at. I've in this in a sense, giving you the answer
right at the beginning because we started with big F, we
differentiated it to find little F, so we knew the answer all the
time. In practice, you won't know the answer all the time, so
one way that we can help ourselves by referring to a
table of antiderivatives. Now a table of antiderivatives will
look something like this one.
There's a copy of this in the notes accompanying the video and
a table of antiderivatives were list lots of functions.
F.
And then the anti derivative.
Big F in the next column and you'll see every anti
derivative will have a plus. See attached to it where C is
an arbitrary constant. Any constant you choose.
What I want to do now is I want to link the concept
of an antiderivative with integration as summation.
Let's think of a function.
Y equals F of X.
Let's suppose the graph of the function looks like this.
I'm going to consider functions which lie entirely
above the X axis, so I'm going to restrict our
attention to the region above the X axis were looking up
here.
And also I want to restrict attention to the right of the Y
axis, so I'm looking to the right of this line here.
Let's ask ourselves what is the area under the graph of Y equals
F of X?
Well, surely that depends on how far I want to move to the
right hand side. Let's suppose I want to move.
To the place where X has an X Coordinate X.
Then clearly, if X is a large
number. The area under this graph here will be large,
whereas if X is a small number, the area under the
graph will be quite small indeed. Effects is actually 0,
so this line is actually lying on the Y axis than the area
under the graph will be 0.
I'm going to do note the area under the graph by a,
so A is the area.
Under
why is F of X?
And as I've just said, the area will depend upon the value that
we choose for X. In other words, A is a function of XA
depends upon X, so we write that like this, a is a of X and
that shows the dependence of the area on the value we choose
for X. As I said, LG X larger area small acts smaller area.
Ask yourself. What is the height of this line here?
Well, this point here.
Lies. On the curve Y equals F of
X. So the Y value at that point is simply the function evaluated
at this X value.
So the Y value here is just F of
X. In other words, the height of this line or the
length of this line is just F of X. The function
evaluated at this point.
Now I'd like you to think what happens if we just increase X by
a very small amount.
So I want to just move this point a little bit further
to the right.
And see what happens.
I'm going to increase X by a little bit and that little bit
that distance in here. I'm going to call Delta X. Delta X stands
for a small change in X or we call it a small increment in X.
By increasing axle little bit.
What we're doing is we're adding a little bit more to this area.
This is the additional contribution to the area. This
shaded region in here.
And that's a little bit extra area, so I'm going to call that.
Delta A, That's an incrementing area changing area.
I'm going to write down an expression for Delta
a try and work it out.
Let's try and see what it is, but I can't get it exactly. But
if I note that the height of this line is F of X.
And the width of this column in here is Delta X. Then I can get
an approximate value for this area by assuming that it's a
rectangular section. In other words, I'm going to ignore this
little bit at the top in there.
If I assume it's a rectangular section, then the area here this
additional area Delta A.
Is F of X multiplied by Delta X.
Now, that's only approximately true, so this is really an
approximately equal to symbol.
This additional area is approximately equal to
F of X Times Delta X.
Let me now divide both sides by Delta X. That's going to give me
Delta a over Delta. X is approximately equal to F of X.
How can we make it more accurate? Well, one way we can
make this more accurate is by choosing this column to be even
thinner by letting Delta X be smaller, because then this
additional contribution that I've got in here that I didn't
count is reducing its reducing in size. So what I want to do is
I want to let Delta X get even smaller and smaller and in the
end I want to take the limit.
As Delta X tends to zero of Delta over Delta X.
And when I do that, this approximation in here will
become exact and that will give us F of X.
Now, if you've studied the unit on differentiation from first
principles, you realize that this in here is the definition
of the derivative of A with respect to X, which we write as
DADX. So we have the result that DADX is a little F of X.
Let's explore this a little bit further.
We have the ADX is a little F of X.
What does this mean? Well, it means that little laugh
is the derivative of A.
But it also means if we think back to the discussion that we
had at the beginning of this video, that a must be the anti
derivative of F. A is an anti derivative.
F. In other words, the area is F of X. The anti derivative of
little F was any constant.
That's going to be an important result. What it's saying is that
if we have a function, why is F of X and we want to find the
area under the graph? What we do is we calculate an anti
derivative of Little F which is
big F. And use this expression.
To find the area.
There's one thing we don't know in this expression at the
moment, and it's this CC. Remember, is an arbitrary
constant, but we can get a value for. See if we just look back
again. So the graph I drew.
And ask yourself what will be the area under this curve when X
is chosen to be zero? Well, if you remember, we said that if
this vertical line here had been on the Y axis.
Then the area under the curve would have been 0.
So this gives us a condition. It tells us that
when X is 0, the area is 0.
When X is 0.
Area is 0. What does this mean? While the area being 0?
X being 0.
Plus a constant and this condition then gives us a value
for C, so C must be equal to minus F of North.
And that value for C and go back in this result here. So we have
the final result that the area under the graph is given by big
F of X minus big F of note.
Let me just write that down again.
OK.
Now let's look at this problem. Supposing that I'm interested in
finding the area under the graph of Y is little F of X.
Up to the point.
Where X has the value be.
So I want the area from the Y axis, which is where we're
working from above the X axis up to the point.
Where X equals B.
We can use this boxed formula here.
When X is be will get a of B. That's the area up to be.
Will be F of B.
Minus F of note.
So this expression will give you the area up to be.
Suppose now I want.
Area up to A and let's Suppose A is about here.
So now I'm interested.
In this area in here, which is the area from the Y axis up to a
well, again using the same formula, the area up to a which
is a of A.
Is F evaluated at the X value which is a?
Subtract. Big F of note.
So I have two expressions on the page here, one for the
area up to be.
And one for the area up to A.
Ask yourself How do I find this area in here? That's
the area between A&B.
Well, the area between A&B we can think of as the area up to
be. Subtract the area up to A.
So if we find the difference of these two quantities, that will
give us the area between A&B.
So we get the area.
Under Wise F of X.
From X equals A to X equals
B. Is the area under this graph from X is A
to X is B is given by?
Once the area up to be, subtract the area up to A.
So if we just find the difference of these two
quantities will have F of B minus F of A.
Add minus F of not minus minus F of notes. These terms will
cancel out. So In other words that's the result we need. The
area under this graph between A&B is just.
Big F of B minus big FA.
Let me just write that down again.
The area and Y equals F of X between ex is an ex is B is
given by big F of B minus big F of a where big F remember is an
anti derivative of Little F.
Any answer derivative?
So this is a very important result because it means that if
I give you a function little F
of X. And you can calculate an anti derivative of its big F.
And you can find the area under the graph merely by
evaluating that anti derivative at B, evaluating
it A and finding the difference of the two
quantities.
Now in the previous video on integration by summation, the
area under under a graph was found in a slightly different
way. What was done there?
Was that the area under the graph between A&B was found by
dividing the area into lots of thin rectangular strips?
Finding the area of each of the rectangles and adding them all
up. And doing that we had this result that the area under the
graph between A&B was given by the limit as Delta X tends to
zero of the sum of all these rectangular areas, which was F
of X Delta X between X is A and X is be. If you're not familiar
with that, I would advise you go back and have a look at that
other video on integration by summation, but that formula.
For this area was derived in that video.
And this formula defines what we mean
by the definite integral from A to B
of F of X DX.
That's how we defined a definite integral.
And if we wanted to define a definite integral, if we wanted
to calculate a definite integral in that video, we had to do it
through the process of finding the limit of a sum.
Now the the process of finding the limit of a sum is quite
cumbersome and impractical. But what we've learned now is that
we don't have to use the limit of a sum to find the area. We
can use these anti derivatives.
Because we've got two expressions for the area, we've
got this expression. As a definite integral, and we've got
this expression in terms of antiderivatives, and if we put
all that together will end up
with this result. The definite integral from A to B.
F of X DX.
Is FB.
Minus F of A.
In other words, the definite integral of little F between the
limits of A&B is found by evaluating this expression,
where big F is an anti derivative of Little F.
Let me write that formula down again.
The integral from A to B.
F of X DX is given by big FB minus big F of A.
Let me give you an example.
Suppose we're interested in the problem of finding
the area under the graph of Y equals X squared.
And let's suppose for the sake of argument we want the area
under the graph. Between X is not an ex is one, so we're
interested in this area.
In the previous unit on integration as a summation,
this was done by dividing this area into lots of thin
rectangular strips.
And finding the area of each of those rectangles separately.
And then adding them all up.
That gave rise to this formula that the area is the limit.
As Delta X tends to 0.
Of the sum from X equals not to X equals 1.
Of X squared Delta X.
So the area was expressed as the limit of a sum.
And in turn, that defines the definite integral. X is not to
one of X squared DX.
Now open till now if you wanted to work this area out
the way you would do it would be by finding the limit of
this some, but that's impractical and cumbersome.
Instead, we're going to use this result using
antiderivatives.
Our little F of X in this case is X squared.
And this formula at the top of the page tells us that we can
evaluate this definite integral by finding an
antiderivative capital F. So we want to do that first.
Well, if little F is X squared, big F of X, well we
want an anti derivative of X squared and if you don't know
what one is, you refer back to your table.
An anti derivative of X squared is X cubed over 3
plus a constant.
So in order to calculate this definite integral, what we need
to do is evaluate.
The Anti Derivative Capital F. At B. That's the upper limit,
which in this case is one.
And then at the lower limit, which in the in our case is
zero. Let's workout F of one.
Well, F of one is going to be 1
cubed. Over 3, which is just a third.
Plus C.
Let's workout big F of 0.
Big F of 0.
Is going to be 0 cubed over three, which is 0 plus. See, so
it's just see. And then we want to find the difference.
Half of 1 minus F of note.
Will be 1/3 plus C minus C, so the Seas will cancel and would
be just left with the third.
In other words, to calculate this integral here.
All we have to do is find the Anti Derivative Capital F.
Evaluate it at the upper limit evaluated at the lower limit.
Find the difference and the result is the third. That's the
area under this graph between North and one.
Now let me just show you how we would normally set this out.
We would normally set this out like this.
We find. An anti derivative of X squared, which we've seen is X
cubed over 3 plus C.
We don't normally write the Plus C down, and the reason for that
is when we're finding definite integrals, the sea will always
cancel out as we saw here, the Seas cancelled out in here and
that will always be the case.
So we don't actually need to write a plus C down when we
write down an anti derivative of X squared.
It's conventional to write down the anti derivative in square
brackets. And to transfer the limits on the original integral,
the Norton one to the right hand side over there like so.
What we then want to do is evaluate the anti derivative at
the top limit that corresponded to the FB or the F of one. So we
work this out at the top limit which is 1 cubed over 3.
We work it out at the bottom limit. That's the F of a or
the F of Norte.
We work this out of the lower limits will just get zero and
then we want the difference between the two.
Which is just going to give us a third. So that's the
normal way we would set out a definite integral.
So as we've seen.
Definite integration is very closely associated with Anti
Differentiation.
Because of this, in general it's useful to think of Anti
Differentiation as integration and then we would often refer to
a table like this one of antiderivatives as simply a
table of integrals.
And there will be a table of integrals in the notes and in
later videos you'll be looking in much more detail at how to
use a table of integrals. So you think of the table of
integrals as your table of antiderivatives and vice versa.
Very early on in the video, we looked at this problem. We
started off with little F of X being equal to four X cubed.
Minus Seven X squared.
Plus 12X minus 4.
We differentiate it.
To give 12 X squared minus 14X plus 12.
And then we said that an anti derivative of 12 X squared minus
14X plus 12.
Was this function capital F plus C?
We've got a notation we can use now because we've
linked antiderivatives to integrals and the notation
we would use is that the integral of 12 X squared.
Minus 14X plus 12 DX is equal to.
4X cubed
minus Seven X squared plus 12X plus any arbitrary constant.
And we call this an indefinite integral.
And we say that the indefinite integral of 12 X squared minus
14X plus 12 with respect to X is 4X cubed minus Seven X squared
plus 12X. Plus a constant of integration.
OK, In summary, what have we found? What we found
that a definite integral?
The integral from A to B of little F of X DX. We found that
this is a number.
And it's obtained from the formula F of B minus FA,
where big F is any anti derivative of Little F.
And we've also seen the indefinite integral of F
of X DX.
Is a function big F of X plus an arbitrary constant? Again, where
F is any anti derivative of Little F&C is an arbitrary
constant. So in this video we've learned how to do
differentiation in reverse. We've learned about
antiderivatives. Definite
integrals. And indefinite integrals in subsequent videos.
You learn a lot more teak techniques of integration.